Yes, there is more bugs... Using Wolfram|Alpha one can find that:
sin'(2x) = cos(2x)
sin'(2*x) = cos(2x)
(sin(2x))' = cos(2x)
but
D[sin(2x),x] = 2 cos(2x)
With other compositions of functions is same:
log'(2x) = 1/2x
while
D[log(2x),x] = 1/x
Above bug is found by user Edo; he write about it on news-group
hr.sci.matematika
I found another bug - W|A has problem with interpreting second-order ODE
like:
y''+y'=cos(x), y(0.5)=0.7, y'(0.7)=0.3
but user gogo222 write on hr.sci.matematika that he found that it can be
solve if we change
y(0.5)=0.7
with
y[0.5]=0.7,
i.e. if we write
y''+y'=cos(x), y[0.5]=0.7, y'(0.7)=0.3
than we get some solution. Is it correct, that is the question. :-)))
Sorry because my english is not so good.
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>Yes, there is more bugs... Using Wolfram|Alpha one can find that:
>sin'(2x) = cos(2x)
>sin'(2*x) = cos(2x)
>(sin(2x))' = cos(2x)
>but
>D[sin(2x),x] = 2 cos(2x)
I see nothing wrong here. The use of ' for derivative
gets the derivative of the function; the use of D gets
the derivative of the functional value with respect to
the specified argument.
D[sin(2x),x] = (sin @ 2*)'(x),
where I have used @ for function composition. The quantity
sin(2x) is (sin @ 2*)(x),
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This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Actually I don't there is any bugs here, in sin'(2x), 2x is the
variable as a whole, but D[sin(2x),x] means sin(2x) is the function,
and x is the variable, that's the definition of D[f(x),x]